Mercurial > sdl-ios-xcode
annotate src/video/e_sqrt.h @ 2661:d38309be5178 gsoc2008_audio_resampling
The windowed sinc filter generation code seems to be working fine. The FIR filtering code is also now working reasonably well. Occasionally the FIR filter will pop, but setting the normalization factor lower seems to help this. I suspect the problem is in the fixed point multiply/add. I also have a hunch the zero stuffing/sample discarding code is not correct, and I'll look at that soon to get it sorted out.
author | Aaron Wishnick <schnarf@gmail.com> |
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date | Wed, 02 Jul 2008 08:04:50 +0000 |
parents | c121d94672cb |
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rev | line source |
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1 /* @(#)e_sqrt.c 5.1 93/09/24 */ |
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2 /* |
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3 * ==================================================== |
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4 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved. |
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5 * |
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6 * Developed at SunPro, a Sun Microsystems, Inc. business. |
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7 * Permission to use, copy, modify, and distribute this |
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8 * software is freely granted, provided that this notice |
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9 * is preserved. |
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10 * ==================================================== |
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11 */ |
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12 |
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13 #if defined(LIBM_SCCS) && !defined(lint) |
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14 static char rcsid[] = "$NetBSD: e_sqrt.c,v 1.8 1995/05/10 20:46:17 jtc Exp $"; |
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15 #endif |
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16 |
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17 /* __ieee754_sqrt(x) |
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18 * Return correctly rounded sqrt. |
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19 * ------------------------------------------ |
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20 * | Use the hardware sqrt if you have one | |
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21 * ------------------------------------------ |
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22 * Method: |
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23 * Bit by bit method using integer arithmetic. (Slow, but portable) |
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24 * 1. Normalization |
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25 * Scale x to y in [1,4) with even powers of 2: |
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26 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then |
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27 * sqrt(x) = 2^k * sqrt(y) |
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28 * 2. Bit by bit computation |
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29 * Let q = sqrt(y) truncated to i bit after binary point (q = 1), |
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30 * i 0 |
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31 * i+1 2 |
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32 * s = 2*q , and y = 2 * ( y - q ). (1) |
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33 * i i i i |
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34 * |
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35 * To compute q from q , one checks whether |
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36 * i+1 i |
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37 * |
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38 * -(i+1) 2 |
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39 * (q + 2 ) <= y. (2) |
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40 * i |
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41 * -(i+1) |
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42 * If (2) is false, then q = q ; otherwise q = q + 2 . |
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43 * i+1 i i+1 i |
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44 * |
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45 * With some algebric manipulation, it is not difficult to see |
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46 * that (2) is equivalent to |
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47 * -(i+1) |
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48 * s + 2 <= y (3) |
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49 * i i |
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50 * |
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51 * The advantage of (3) is that s and y can be computed by |
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52 * i i |
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53 * the following recurrence formula: |
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54 * if (3) is false |
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55 * |
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56 * s = s , y = y ; (4) |
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57 * i+1 i i+1 i |
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58 * |
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59 * otherwise, |
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60 * -i -(i+1) |
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61 * s = s + 2 , y = y - s - 2 (5) |
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62 * i+1 i i+1 i i |
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63 * |
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64 * One may easily use induction to prove (4) and (5). |
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65 * Note. Since the left hand side of (3) contain only i+2 bits, |
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66 * it does not necessary to do a full (53-bit) comparison |
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67 * in (3). |
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68 * 3. Final rounding |
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69 * After generating the 53 bits result, we compute one more bit. |
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70 * Together with the remainder, we can decide whether the |
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71 * result is exact, bigger than 1/2ulp, or less than 1/2ulp |
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72 * (it will never equal to 1/2ulp). |
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73 * The rounding mode can be detected by checking whether |
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74 * huge + tiny is equal to huge, and whether huge - tiny is |
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75 * equal to huge for some floating point number "huge" and "tiny". |
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76 * |
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77 * Special cases: |
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78 * sqrt(+-0) = +-0 ... exact |
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79 * sqrt(inf) = inf |
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80 * sqrt(-ve) = NaN ... with invalid signal |
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81 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN |
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82 * |
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83 * Other methods : see the appended file at the end of the program below. |
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84 *--------------- |
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85 */ |
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86 |
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87 /*#include "math.h"*/ |
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88 #include "math_private.h" |
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89 |
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90 #ifdef __STDC__ |
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91 double SDL_NAME(copysign) (double x, double y) |
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92 #else |
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93 double SDL_NAME(copysign) (x, y) |
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94 double |
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95 x, |
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96 y; |
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97 #endif |
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98 { |
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99 u_int32_t hx, hy; |
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100 GET_HIGH_WORD(hx, x); |
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101 GET_HIGH_WORD(hy, y); |
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102 SET_HIGH_WORD(x, (hx & 0x7fffffff) | (hy & 0x80000000)); |
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103 return x; |
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104 } |
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105 |
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106 #ifdef __STDC__ |
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107 double SDL_NAME(scalbn) (double x, int n) |
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108 #else |
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109 double SDL_NAME(scalbn) (x, n) |
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110 double |
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111 x; |
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112 int |
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113 n; |
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114 #endif |
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115 { |
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116 int32_t k, hx, lx; |
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117 EXTRACT_WORDS(hx, lx, x); |
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118 k = (hx & 0x7ff00000) >> 20; /* extract exponent */ |
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119 if (k == 0) { /* 0 or subnormal x */ |
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120 if ((lx | (hx & 0x7fffffff)) == 0) |
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121 return x; /* +-0 */ |
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122 x *= two54; |
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123 GET_HIGH_WORD(hx, x); |
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124 k = ((hx & 0x7ff00000) >> 20) - 54; |
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125 if (n < -50000) |
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126 return tiny * x; /*underflow */ |
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127 } |
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128 if (k == 0x7ff) |
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129 return x + x; /* NaN or Inf */ |
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130 k = k + n; |
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131 if (k > 0x7fe) |
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132 return huge * SDL_NAME(copysign) (huge, x); /* overflow */ |
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133 if (k > 0) { /* normal result */ |
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134 SET_HIGH_WORD(x, (hx & 0x800fffff) | (k << 20)); |
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135 return x; |
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136 } |
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137 if (k <= -54) { |
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138 if (n > 50000) /* in case integer overflow in n+k */ |
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139 return huge * SDL_NAME(copysign) (huge, x); /*overflow */ |
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140 else |
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141 return tiny * SDL_NAME(copysign) (tiny, x); /*underflow */ |
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142 } |
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143 k += 54; /* subnormal result */ |
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144 SET_HIGH_WORD(x, (hx & 0x800fffff) | (k << 20)); |
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145 return x * twom54; |
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146 } |
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147 |
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148 #ifdef __STDC__ |
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149 double |
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150 __ieee754_sqrt(double x) |
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151 #else |
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152 double |
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153 __ieee754_sqrt(x) |
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154 double x; |
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155 #endif |
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156 { |
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157 double z; |
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158 int32_t sign = (int) 0x80000000; |
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159 int32_t ix0, s0, q, m, t, i; |
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160 u_int32_t r, t1, s1, ix1, q1; |
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161 |
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162 EXTRACT_WORDS(ix0, ix1, x); |
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163 |
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164 /* take care of Inf and NaN */ |
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165 if ((ix0 & 0x7ff00000) == 0x7ff00000) { |
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166 return x * x + x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf |
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167 sqrt(-inf)=sNaN */ |
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168 } |
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169 /* take care of zero */ |
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170 if (ix0 <= 0) { |
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171 if (((ix0 & (~sign)) | ix1) == 0) |
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172 return x; /* sqrt(+-0) = +-0 */ |
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173 else if (ix0 < 0) |
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174 return (x - x) / (x - x); /* sqrt(-ve) = sNaN */ |
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175 } |
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176 /* normalize x */ |
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177 m = (ix0 >> 20); |
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178 if (m == 0) { /* subnormal x */ |
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179 while (ix0 == 0) { |
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180 m -= 21; |
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181 ix0 |= (ix1 >> 11); |
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182 ix1 <<= 21; |
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183 } |
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184 for (i = 0; (ix0 & 0x00100000) == 0; i++) |
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185 ix0 <<= 1; |
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186 m -= i - 1; |
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187 ix0 |= (ix1 >> (32 - i)); |
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188 ix1 <<= i; |
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189 } |
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190 m -= 1023; /* unbias exponent */ |
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191 ix0 = (ix0 & 0x000fffff) | 0x00100000; |
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192 if (m & 1) { /* odd m, double x to make it even */ |
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193 ix0 += ix0 + ((ix1 & sign) >> 31); |
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194 ix1 += ix1; |
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195 } |
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196 m >>= 1; /* m = [m/2] */ |
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197 |
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198 /* generate sqrt(x) bit by bit */ |
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199 ix0 += ix0 + ((ix1 & sign) >> 31); |
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200 ix1 += ix1; |
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201 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */ |
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202 r = 0x00200000; /* r = moving bit from right to left */ |
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203 |
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204 while (r != 0) { |
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205 t = s0 + r; |
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206 if (t <= ix0) { |
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207 s0 = t + r; |
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208 ix0 -= t; |
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209 q += r; |
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210 } |
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211 ix0 += ix0 + ((ix1 & sign) >> 31); |
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SDL 1.2 is moving to a branch, and SDL 1.3 is becoming the head.
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212 ix1 += ix1; |
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213 r >>= 1; |
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214 } |
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215 |
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216 r = sign; |
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217 while (r != 0) { |
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218 t1 = s1 + r; |
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219 t = s0; |
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220 if ((t < ix0) || ((t == ix0) && (t1 <= ix1))) { |
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221 s1 = t1 + r; |
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222 if (((int32_t) (t1 & sign) == sign) && (s1 & sign) == 0) |
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223 s0 += 1; |
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224 ix0 -= t; |
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225 if (ix1 < t1) |
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226 ix0 -= 1; |
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227 ix1 -= t1; |
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228 q1 += r; |
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229 } |
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230 ix0 += ix0 + ((ix1 & sign) >> 31); |
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231 ix1 += ix1; |
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232 r >>= 1; |
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233 } |
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234 |
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235 /* use floating add to find out rounding direction */ |
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236 if ((ix0 | ix1) != 0) { |
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237 z = one - tiny; /* trigger inexact flag */ |
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238 if (z >= one) { |
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239 z = one + tiny; |
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240 if (q1 == (u_int32_t) 0xffffffff) { |
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241 q1 = 0; |
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242 q += 1; |
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243 } else if (z > one) { |
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244 if (q1 == (u_int32_t) 0xfffffffe) |
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245 q += 1; |
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246 q1 += 2; |
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247 } else |
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248 q1 += (q1 & 1); |
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249 } |
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250 } |
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251 ix0 = (q >> 1) + 0x3fe00000; |
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252 ix1 = q1 >> 1; |
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253 if ((q & 1) == 1) |
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254 ix1 |= sign; |
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255 ix0 += (m << 20); |
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256 INSERT_WORDS(z, ix0, ix1); |
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257 return z; |
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258 } |
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259 |
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260 /* |
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261 Other methods (use floating-point arithmetic) |
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262 ------------- |
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263 (This is a copy of a drafted paper by Prof W. Kahan |
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264 and K.C. Ng, written in May, 1986) |
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265 |
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266 Two algorithms are given here to implement sqrt(x) |
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267 (IEEE double precision arithmetic) in software. |
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268 Both supply sqrt(x) correctly rounded. The first algorithm (in |
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269 Section A) uses newton iterations and involves four divisions. |
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270 The second one uses reciproot iterations to avoid division, but |
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271 requires more multiplications. Both algorithms need the ability |
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272 to chop results of arithmetic operations instead of round them, |
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273 and the INEXACT flag to indicate when an arithmetic operation |
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274 is executed exactly with no roundoff error, all part of the |
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275 standard (IEEE 754-1985). The ability to perform shift, add, |
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276 subtract and logical AND operations upon 32-bit words is needed |
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277 too, though not part of the standard. |
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278 |
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279 A. sqrt(x) by Newton Iteration |
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280 |
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281 (1) Initial approximation |
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282 |
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283 Let x0 and x1 be the leading and the trailing 32-bit words of |
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284 a floating point number x (in IEEE double format) respectively |
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285 |
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286 1 11 52 ...widths |
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287 ------------------------------------------------------ |
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288 x: |s| e | f | |
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289 ------------------------------------------------------ |
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290 msb lsb msb lsb ...order |
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291 |
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292 |
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293 ------------------------ ------------------------ |
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294 x0: |s| e | f1 | x1: | f2 | |
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295 ------------------------ ------------------------ |
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296 |
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297 By performing shifts and subtracts on x0 and x1 (both regarded |
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298 as integers), we obtain an 8-bit approximation of sqrt(x) as |
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299 follows. |
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300 |
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301 k := (x0>>1) + 0x1ff80000; |
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302 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits |
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303 Here k is a 32-bit integer and T1[] is an integer array containing |
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304 correction terms. Now magically the floating value of y (y's |
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305 leading 32-bit word is y0, the value of its trailing word is 0) |
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306 approximates sqrt(x) to almost 8-bit. |
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307 |
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308 Value of T1: |
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309 static int T1[32]= { |
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310 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592, |
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311 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215, |
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312 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581, |
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313 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,}; |
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314 |
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315 (2) Iterative refinement |
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316 |
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317 Apply Heron's rule three times to y, we have y approximates |
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318 sqrt(x) to within 1 ulp (Unit in the Last Place): |
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319 |
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320 y := (y+x/y)/2 ... almost 17 sig. bits |
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321 y := (y+x/y)/2 ... almost 35 sig. bits |
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322 y := y-(y-x/y)/2 ... within 1 ulp |
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323 |
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324 |
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325 Remark 1. |
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326 Another way to improve y to within 1 ulp is: |
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327 |
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328 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x) |
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329 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x) |
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330 |
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331 2 |
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332 (x-y )*y |
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333 y := y + 2* ---------- ...within 1 ulp |
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334 2 |
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335 3y + x |
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336 |
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337 |
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338 This formula has one division fewer than the one above; however, |
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339 it requires more multiplications and additions. Also x must be |
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340 scaled in advance to avoid spurious overflow in evaluating the |
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341 expression 3y*y+x. Hence it is not recommended uless division |
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342 is slow. If division is very slow, then one should use the |
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343 reciproot algorithm given in section B. |
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344 |
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345 (3) Final adjustment |
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346 |
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347 By twiddling y's last bit it is possible to force y to be |
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348 correctly rounded according to the prevailing rounding mode |
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349 as follows. Let r and i be copies of the rounding mode and |
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350 inexact flag before entering the square root program. Also we |
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351 use the expression y+-ulp for the next representable floating |
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352 numbers (up and down) of y. Note that y+-ulp = either fixed |
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353 point y+-1, or multiply y by nextafter(1,+-inf) in chopped |
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354 mode. |
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355 |
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356 I := FALSE; ... reset INEXACT flag I |
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357 R := RZ; ... set rounding mode to round-toward-zero |
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358 z := x/y; ... chopped quotient, possibly inexact |
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359 If(not I) then { ... if the quotient is exact |
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360 if(z=y) { |
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361 I := i; ... restore inexact flag |
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362 R := r; ... restore rounded mode |
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363 return sqrt(x):=y. |
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364 } else { |
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365 z := z - ulp; ... special rounding |
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366 } |
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367 } |
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368 i := TRUE; ... sqrt(x) is inexact |
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369 If (r=RN) then z=z+ulp ... rounded-to-nearest |
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370 If (r=RP) then { ... round-toward-+inf |
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371 y = y+ulp; z=z+ulp; |
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372 } |
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373 y := y+z; ... chopped sum |
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374 y0:=y0-0x00100000; ... y := y/2 is correctly rounded. |
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375 I := i; ... restore inexact flag |
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376 R := r; ... restore rounded mode |
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377 return sqrt(x):=y. |
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378 |
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379 (4) Special cases |
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380 |
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381 Square root of +inf, +-0, or NaN is itself; |
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382 Square root of a negative number is NaN with invalid signal. |
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383 |
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384 |
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385 B. sqrt(x) by Reciproot Iteration |
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386 |
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387 (1) Initial approximation |
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388 |
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389 Let x0 and x1 be the leading and the trailing 32-bit words of |
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390 a floating point number x (in IEEE double format) respectively |
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391 (see section A). By performing shifs and subtracts on x0 and y0, |
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392 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows. |
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393 |
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394 k := 0x5fe80000 - (x0>>1); |
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395 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits |
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396 |
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397 Here k is a 32-bit integer and T2[] is an integer array |
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398 containing correction terms. Now magically the floating |
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399 value of y (y's leading 32-bit word is y0, the value of |
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400 its trailing word y1 is set to zero) approximates 1/sqrt(x) |
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401 to almost 7.8-bit. |
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402 |
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403 Value of T2: |
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404 static int T2[64]= { |
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405 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866, |
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406 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f, |
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407 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d, |
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408 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0, |
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409 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989, |
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410 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd, |
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411 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e, |
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412 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,}; |
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413 |
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414 (2) Iterative refinement |
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415 |
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416 Apply Reciproot iteration three times to y and multiply the |
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417 result by x to get an approximation z that matches sqrt(x) |
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418 to about 1 ulp. To be exact, we will have |
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419 -1ulp < sqrt(x)-z<1.0625ulp. |
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420 |
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421 ... set rounding mode to Round-to-nearest |
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422 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x) |
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423 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x) |
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424 ... special arrangement for better accuracy |
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425 z := x*y ... 29 bits to sqrt(x), with z*y<1 |
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426 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x) |
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427 |
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428 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that |
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429 (a) the term z*y in the final iteration is always less than 1; |
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430 (b) the error in the final result is biased upward so that |
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431 -1 ulp < sqrt(x) - z < 1.0625 ulp |
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432 instead of |sqrt(x)-z|<1.03125ulp. |
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433 |
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434 (3) Final adjustment |
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435 |
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436 By twiddling y's last bit it is possible to force y to be |
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437 correctly rounded according to the prevailing rounding mode |
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438 as follows. Let r and i be copies of the rounding mode and |
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439 inexact flag before entering the square root program. Also we |
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440 use the expression y+-ulp for the next representable floating |
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441 numbers (up and down) of y. Note that y+-ulp = either fixed |
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442 point y+-1, or multiply y by nextafter(1,+-inf) in chopped |
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443 mode. |
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444 |
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445 R := RZ; ... set rounding mode to round-toward-zero |
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446 switch(r) { |
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447 case RN: ... round-to-nearest |
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448 if(x<= z*(z-ulp)...chopped) z = z - ulp; else |
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449 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp; |
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450 break; |
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451 case RZ:case RM: ... round-to-zero or round-to--inf |
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452 R:=RP; ... reset rounding mod to round-to-+inf |
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453 if(x<z*z ... rounded up) z = z - ulp; else |
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454 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp; |
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455 break; |
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456 case RP: ... round-to-+inf |
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457 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else |
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458 if(x>z*z ...chopped) z = z+ulp; |
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459 break; |
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460 } |
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461 |
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462 Remark 3. The above comparisons can be done in fixed point. For |
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463 example, to compare x and w=z*z chopped, it suffices to compare |
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464 x1 and w1 (the trailing parts of x and w), regarding them as |
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465 two's complement integers. |
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466 |
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467 ...Is z an exact square root? |
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468 To determine whether z is an exact square root of x, let z1 be the |
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469 trailing part of z, and also let x0 and x1 be the leading and |
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470 trailing parts of x. |
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471 |
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472 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0 |
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473 I := 1; ... Raise Inexact flag: z is not exact |
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474 else { |
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475 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2 |
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476 k := z1 >> 26; ... get z's 25-th and 26-th |
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477 fraction bits |
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478 I := i or (k&j) or ((k&(j+j+1))!=(x1&3)); |
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479 } |
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480 R:= r ... restore rounded mode |
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481 return sqrt(x):=z. |
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482 |
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483 If multiplication is cheaper then the foregoing red tape, the |
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484 Inexact flag can be evaluated by |
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485 |
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486 I := i; |
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487 I := (z*z!=x) or I. |
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488 |
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489 Note that z*z can overwrite I; this value must be sensed if it is |
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490 True. |
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491 |
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492 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be |
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493 zero. |
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494 |
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495 -------------------- |
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496 z1: | f2 | |
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497 -------------------- |
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498 bit 31 bit 0 |
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499 |
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500 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd |
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501 or even of logb(x) have the following relations: |
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502 |
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503 ------------------------------------------------- |
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504 bit 27,26 of z1 bit 1,0 of x1 logb(x) |
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505 ------------------------------------------------- |
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506 00 00 odd and even |
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507 01 01 even |
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508 10 10 odd |
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509 10 00 even |
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510 11 01 even |
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511 ------------------------------------------------- |
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512 |
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513 (4) Special cases (see (4) of Section A). |
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514 |
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515 */ |
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516 /* vi: set ts=4 sw=4 expandtab: */ |